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How do you differentiate #f(t)=root3(1+tant)#?

1 Answer

Feb 23, 2017

#f'(t) = sec^2t/(3(1+tant)^(2/3))#

Explanation:

#f(t) = root3(1+tant)#

#= (1+tant)^(1/3)#

#f'(t) = 1/3(1+tant)^(-2/3) * d/dt(1+tant)# [Power rule and Chain rule]

#= 1/3(1+tant)^(-2/3) * (0+ sec^2t)# [Standard differential]

#=sec^2t/(3(1+tant)^(2/3))#

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